Answer
The value of the expression $2{{x}^{3}}+3{{x}^{2}}-8x+3=0$ is $\left\{ -3,\frac{1}{2},1 \right\}$
Work Step by Step
We have to find the value of x from $2{{x}^{3}}+3{{x}^{2}}-8x+3=0$
By the trial method, $ x=1$ is a solution of the equation.
So, $ x-1$ is a factor of the given equation.
Then, divide $2{{x}^{3}}+3{{x}^{2}}-8x+3=0$ by $ x-1$:
$\frac{\left( 2{{x}^{3}}+3{{x}^{2}}-8x+3 \right)}{x-1}=2{{x}^{2}}+5x-3$
We have obtained a quadratic equation. Now, obtain the factors of $2{{x}^{2}}+5x-3$ as given below.
$\begin{align}
& 2{{x}^{2}}+5x-3=0 \\
& 2{{x}^{2}}+6x-x-3=0 \\
& 2x\left( x+3 \right)-1\left( x+3 \right)=0 \\
& \left( 2x-1 \right)\left( x+3 \right)=0
\end{align}$
So, $ x=\frac{1}{2}$ and $ x=-3$.
Hence, the values of x are $\left\{ -3,\frac{1}{2},1 \right\}$.
